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Chapter
9
DIFFERENTIAL EQUATIONS
9.1 Overview
(i)
An equation involving derivative (derivatives) of the dependent variable with
respect to independent variable (variables) is called a differential equation.
(ii)
A differential equation involving derivatives of the dependent variable with
respect to only one independent variable is called an ordinary differential
equation and a differential equation involving derivatives with respect to more
than one independent variables is called a partial differential equation.
(iii)
Order of a differential equation is the order of the highest order derivative
occurring in the differential equation.
(iv)
Degree of a differential equation is defined if it is a polynomial equation in its
derivatives.
(v)
Degree (when defined) of a differential equation is the highest power (positive
integer only) of the highest order derivative in it.
(vi)
A relation between involved variables, which satisfy the given differential
equation is called its solution. The solution which contains as many arbitrary
constants as the order of the differential equation is called the general solution
and the solution free from arbitrary constants is called particular solution.
(vii)
To form a differential equation from a given function, we differentiate the
function successively as many times as the number of arbitrary constants in the
given function and then eliminate the arbitrary constants.
(viii) The order of a differential equation representing a family of curves is same as
the number of arbitrary constants present in the equation corresponding to the
family of curves.
(ix) ‘Variable separable method’ is used to solve such an equation in which variables
can be separated completely, i.e., terms containing x should remain with dx and
terms containing y should remain with dy.
180 MATHEMATICS
(x)
A function F (x, y) is said to be a homogeneous function of degree n if
F (λx, λy )= λn F (x, y) for some non-zero constant λ.
(xi)
A differential equation which can be expressed in the form
dy
= F (x, y) or
dx
dx
= G (x, y), where F (x, y) and G (x, y) are homogeneous functions of degree
dy
zero, is called a homogeneous differential equation.
dy
= F (x, y), we make
dx
substitution y = vx and to solve a homogeneous differential equation of the type
(xii) To solve a homogeneous differential equation of the type
dx
= G (x, y), we make substitution x = vy.
dy
dy
+ Py = Q, where P and Q are constants or
dx
functions of x only is known as a first order linear differential equation. Solution
(xiii) A differential equation of the form
of such a differential equation is given by y (I.F.) =
∫ ( Q × I.F.) dx + C, where
Pdx
I.F. (Integrating Factor) = e ∫ .
(xiv) Another form of first order linear differential equation is
dx
+ P1x = Q1, where
dy
P1 and Q1 are constants or functions of y only. Solution of such a differential
equation is given by x (I.F.) =
∫ ( Q1 × I.F.) dy + C, where I.F. = e∫ P dy .
1
9.2 Solved Examples
Short Answer (S.A.)
Example 1 Find the differential equation of the family of curves y = Ae2x + B.e–2x.
Solution y = Ae2x + B.e–2x
DIFFERENTIAL EQUATIONS 181
dy
= 2Ae2x – 2 B.e–2x and
dx
d2y
= 4Ae2x + 4Be–2x
dx 2
d2y
d2y
=
4y
i.e.,
– 4y = 0.
dx 2
dx 2
Thus
Example 2 Find the general solution of the differential equation
dy y
=
dx x
Solution
⇒
dy
dx
=
y
x
⇒
dy
=
y
dy y
= .
dx x
dx
x
⇒ logy = logx + logc ⇒ y = cx
Example 3 Given that
Solution
dy
= yex ⇒
dx
dy
= yex and x = 0, y = e. Find the value of y when x = 1.
dx
dy
y
=
⇒
e x dx
logy = ex + c
Substituting x = 0 and y = e,we get loge = e0 + c, i.e., c = 0 (∵ loge = 1)
Therefore, log y = ex.
Now, substituting x = 1 in the above, we get log y = e ⇒ y = ee.
Example 4 Solve the differential equation
Solution The equation is of the type
dy
y
+ = x2.
dx
x
dy
+ Py = Q , which is a linear differential
dx
equation.
1
Now I.F. =
∫ x dx = e
logx
= x.
Therefore, solution of the given differential equation is
182 MATHEMATICS
x x 2 dx , i.e. yx =
y.x =
x3
4
Hence y =
x4
4
c
c
.
x
Example 5 Find the differential equation of the family of lines through the origin.
Solution Let y = mx be the family of lines through origin. Therefore,
dy
=m
dx
dy
dy
. x or x
– y = 0.
dx
dx
Example 6 Find the differential equation of all non-horizontal lines in a plane.
Solution The general equation of all non-horizontal lines in a plane is
ax + by = c, where a ≠ 0.
Eliminating m, we get y =
Therefore, a
dx
b = 0.
dy
Again, differentiating both sides w.r.t. y, we get
a
d2x
d 2x
=
0
⇒
= 0.
dy 2
dy 2
Example 7 Find the equation of a curve whose tangent at any point on it, different
from origin, has slope y
Solution Given
⇒
dy
y
dy
dx
1
y
y
.
x
y
x
= y 1
1
x
1
dx
x
Integrating both sides, we get
logy = x + logx + c ⇒
log
y
=x+c
x
DIFFERENTIAL EQUATIONS 183
⇒
y
= ex + c = ex.ec ⇒
x
y
= k . ex
x
⇒ y = kx . ex.
Long Answer (L.A.)
Example 8 Find the equation of a curve passing through the point (1, 1) if the
perpendicular distance of the origin from the normal at any point P(x, y) of the curve
is equal to the distance of P from the x – axis.
– dx
Solution Let the equation of normal at P(x, y) be Y – y = dy ( X – x ) ,i.e.,
Y+ X
dx
–
dy
y
x
dx
dy
=0
...(1)
Therefore, the length of perpendicular from origin to (1) is
y x
1
dx
dy
dx
dy
...(2)
2
Also distance between P and x-axis is |y|. Thus, we get
y x
dx
dy
dx
dy
1
2
2
= |y|
⎛
dx ⎞
2
⇒ ⎜ y+x ⎟ = y 1
dy ⎠
⎝
or
dx
dy
2 xy
dx
= 2 2
y –x
dy
2
⇒
dx dx 2
x – y2
dy dy
2 xy
0 ⇒
dx
0
dy
184 MATHEMATICS
dx
= 0 ⇒ dx = 0
dy
Case I:
Integrating both sides, we get x = k, Substituting x = 1, we get k = 1.
Therefore, x = 1 is the equation of curve (not possible, so rejected).
2x y
dx
= 2 2
y x
dy
Case II:
v x
=
y 2 x2
. Substituting y = vx, we get
2 xy
dy
dx
dv v 2 x 2 x 2
dx
2vx 2
−(1 + v 2 )
2v
⇒ x.
⇒
dv v 2 1
v
2v
dx
2v
dv
1 v2
dx
x
Integrating both sides, we get
log (1 + v2) = – logx + logc
⇒
⇒ log (1 + v2) (x) = log c ⇒ (1 + v2) x = c
x2 + y2 = cx. Substituting x = 1,
Therefore,
y = 1,
we get c = 2.
x2 + y2 – 2x = 0 is the required equation.
Example 9 Find the equation of a curve passing through 1,
tangent to the curve at any point P (x, y) is
4
if the slope of the
y
y
− cos 2 .
x
x
Solution According to the given condition
dy y
y
= − cos 2
dx x
x
... (i)
This is a homogeneous differential equation. Substituting y = vx, we get
v+x
dv
= v – cos2v
dx
⇒
x
dv
= – cos2v
dx
DIFFERENTIAL EQUATIONS 185
⇒
sec2v dv = −
⇒
tan
dx
x
⇒
tan v = – logx + c
y
+ log x = c
x
Substituting x = 1, y =
y
x
tan
...(ii)
4
, we get. c = 1. Thus, we get
+ log x = 1, which is the required equation.
2
Example 10 Solve x
dy
⎛ y⎞
π
xy = 1 + cos ⎜ ⎟ , x ≠ 0 and x = 1, y =
dx
2
⎝x⎠
Solution Given equation can be written as
x2
dy
⎛ y ⎞
xy = 2cos2 ⎜ ⎟ , x ≠ 0.
dx
⎝ 2x ⎠
dy
xy
dx
y
⇒
2cos 2
2x
x2
1
⇒
y
2x
sec 2
2
x2
dy
xy
dx
1
Dividing both sides by x3 , we get
⎛ y ⎞
sec 2 ⎜ ⎟ ⎡ x dy − y ⎤
⎥ 1
⎝ 2 x ⎠ ⎢ dx
⎢
⎥= 3
2
2
⎢ x
⎥ x
⎣
⎦
Integrating both sides, we get
tan
y
2x
1
2 x2
k.
⇒
d
y
tan
dx
2x
1
x3
186 MATHEMATICS
Substituting x = 1, y =
k=
2
, we get
y
3
, therefore, tan
2x
2
1
2 x2
3
is the required solution.
2
Example 11 State the type of the differential equation for the equation.
xdy – ydx =
x2
y 2 dx and solve it.
x2
Solution Given equation can be written as xdy =
x2
dy
dx
y2
x
y2
y dx , i.e.,
y
... (1)
Clearly RHS of (1) is a homogeneous function of degree zero. Therefore, the given
equation is a homogeneous differential equation. Substituting y = vx, we get from (1)
v x
x
dv
dx
dv
dx
x2 v2 x2
x
1 v2
⇒
vx
dv
1 v
2
i.e. v x
dv
dx
1 v2 v
dx
x
... (2)
Integrating both sides of (2), we get
log (v + 1 v 2 ) = logx + logc ⇒ v + 1 v 2 = cx
⇒
y
y2
+ 1 2 = cx
x
x
⇒ y+
x2
y 2 = cx2
DIFFERENTIAL EQUATIONS 187
Objective Type Questions
Choose the correct answer from the given four options in each of the Examples 12 to 21.
2
3
dy ⎞ ⎛ d 2 y ⎞
⎛
Example 12 The degree of the differential equation ⎜1 +
⎟ is
⎟ =⎜
dx ⎠ ⎝ dx 2 ⎠
⎝
(A) 1
(B) 2
(C) 3
(D) 4
Solution The correct answer is (B).
Example 13 The degree of the differential equation
2
⎛ d2y ⎞
d2y
⎛ dy ⎞
2
+
3
=
x
log
⎜ 2 ⎟ is
⎜ ⎟
dx 2
⎝ dx ⎠
⎝ dx ⎠
(A) 1
(B) 2
(C) 3
(D) not defined
Solution Correct answer is (D). The given differential equation is not a polynomial
equation in terms of its derivatives, so its degree is not defined.
2
⎡ ⎛ dy ⎞ 2 ⎤ d 2 y
Example 14 The order and degree of the differential equation ⎢1+ ⎜ ⎟ ⎥ = 2
⎢⎣ ⎝ dx ⎠ ⎥⎦ dx
respectively, are
(A) 1, 2
(B) 2, 2
(C) 2, 1
(D) 4, 2
Solution Correct answer is (C).
Example 15 The order of the differential equation of all circles of given radius a is:
(A) 1
(B) 2
(C) 3
(D) 4
Solution Correct answer is (B). Let the equation of given family be
(x – h)2 + (y – k)2 = a2 . It has two orbitrary constants h and k. Threrefore, the order of
the given differential equation will be 2.
Example 16 The solution of the differential equation 2 x .
(A) straight lines (B) circles
dy
– y = 3 represents a family of
dx
(C) parabolas
(D) ellipses
188 MATHEMATICS
Solution Correct answer is (C). Given equation can be written as
2dy dx
y 3 x ⇒ 2log (y + 3) = logx + logc
⇒
(y + 3)2 = cx which represents the family of parabolas
Example 17 The integrating factor of the differential equation
dy
(x log x) + y = 2logx is
dx
(A) ex
(B) log x
(C) log (log x)
(D) x
dy
Solution Correct answer is (B). Given equation can be written as dx
y
2
x log x x .
1
Therefore,
dx
I.F. = ∫ x log x
= elog (logx)
e
= log x.
Example 18 A solution of the differential equation
(A) y = 2
(B) y = 2x
dy
dx
2
x
dy
dx
y
0 is
(D) y = 2x2 – 4
(C) y = 2x – 4
Solution Correct answer is (C).
Example 19 Which of the following is not a homogeneous function of x and y.
(A) x2 + 2xy
(B) 2x – y
2
(C) cos
y
x
y
x
(D) sinx – cosy
Solution Correct answer is (D).
dx dy
Example 20 Solution of the differential equation x + y = 0 is
1 1
(A) x + y = c
(B) logx . logy = c
(C) xy = c
(D) x + y = c
Solution Correct answer is (C). From the given equation, we get logx + logy = logc
giving xy = c.
DIFFERENTIAL EQUATIONS 189
Example 21 The solution of the differential equation x
(A) y =
x2 + c
4 x2
(B) y =
Solution Correct answer is (D). I.F. = e
is y . x2 =
2
∫ x .xdx =
2
dx
x
x2
+c
4
dy
2 y x 2 is
dx
(C) y =
e2log x elog x
2
x4 + c
x4 + c
y
=
(D)
x2
4 x2
x 2 . Therefore, the solution
x4 c
x4
+ k , i.e., y =
.
4 x2
4
Example 22 Fill in the blanks of the following:
(i)
Order of the differential equation representing the family of parabolas
y2 = 4ax is __________ .
2
(ii)
3
2
⎛ dy ⎞ ⎛ d y ⎞
+
The degree of the differential equation ⎜ ⎟ ⎜ 2 ⎟ = 0 is ________ .
⎝ dx ⎠ ⎝ dx ⎠
(iii)
The number of arbitrary constants in a particular solution of the differential
equation tan x dx + tan y dy = 0 is __________ .
(iv)
F (x, y) =
(v)
x2 + y 2 + y
is a homogeneous function of degree__________ .
x
An appropriate substitution to solve the differential equation
x 2 log
dx
=
dy
x
y
xy log
x2
x
y
is__________ .
dy
− y = sinx is __________ .
dx
(vi)
Integrating factor of the differential equation x
(vii)
The general solution of the differential equation
dy
= e x − y is __________ .
dx
190 MATHEMATICS
(viii)
(ix)
dy y
+ =1 is __________ .
dx x
The differential equation representing the family of curves y = A sinx + B
cosx is __________ .
The general solution of the differential equation
e
2 x
(x)
x
y dx
1( x 0) when written in the form dy + Py = Q , then
x dy
dx
P = __________ .
Solution
(i)
(ii)
(iii)
(iv)
(v)
One; a is the only arbitrary constant.
Two; since the degree of the highest order derivative is two.
Zero; any particular solution of a differential equation has no arbitrary constant.
Zero.
x = vy.
(vi)
1
dy y sin x
− =
; given differential equation can be written as
and therefore
dx x
x
x
(vii)
1
.
x
ey = ex + c from given equation, we have eydy = exdx.
(viii)
xy =
(ix)
d2y
+ y = 0; Differentiating the given function w.r.t. x successively, we get
dx 2
I.F. = e
x2
2
1
dx
x
= e–logx =
c ; I.F. =
e
dy
= Acosx – Bsinx
dx
= elogx = x and the solution is y . x = x .1 dx =
and
d2y
= –Asinx – Bcosx
dx 2
d2y
+ y = 0 is the differential equation.
dx 2
⇒
(x)
1
dx
x
1
x
; the given equation can be written as
x2
+C.
2
DIFFERENTIAL EQUATIONS 191
dy
e –2 x
=
dx
x
y
x
i.e.
y
dy
e –2 x
+
=
x
dx
x
dy
+ Py = Q.
dx
Example 23 State whether the following statements are True or False.
(i)
Order of the differential equation representing the family of ellipses having
centre at origin and foci on x-axis is two.
This is a differential equation of the type
(ii)
(iii)
Degree of the differential equation 1+
dy
d2y
is not defined.
2 =x+
dx
dx
dy
dy
y 5 is a differential equation of the type
+ Py = Q but it can be solved
dx
dx
using variable separable method also.
⎛
y cos ⎜
⎝
y⎞
⎟+ x
x⎠
⎛ y ⎞ is not a homogeneous function.
x cos ⎜ ⎟
⎝x⎠
(iv)
F(x, y) =
(v)
x2 y 2
F(x, y) =
is a homogeneous function of degree 1.
x y
dy
dx
y cos x is ex.
(vi)
Integrating factor of the differential equation
(vii)
The general solution of the differential equation x(1 + y2)dx + y (1 + x2)dy = 0
is (1 + x2) (1 + y2) = k.
(viii)
The general solution of the differential equation
dy
+ y sec x = tanx is
dx
y (secx – tanx) = secx – tanx + x + k.
(ix)
x + y = tan–1y is a solution of the differential equation y2
dy
y2 1 0
dx
192 MATHEMATICS
(x)
y = x is a particular solution of the differential equation
d 2 y 2 dy
x
dx
dx 2
xy x .
Solution
(i)
True, since the equation representing the given family is
x2
a2
y2
1 , which
b2
has two arbitrary constants.
(ii)
True, because it is not a polynomial equation in its derivatives.
(iii)
True
(iv)
True, because f ( λx, λy) = λ° f (x, y).
(v)
True, because f ( λx, λy) = λ1 f (x, y).
(vi)
False, because I.F = e
(vii)
True, because given equation can be written as
1dx
e– x .
2x
2y
dx
dy
2
1 x
1 y2
(viii)
⇒
log (1 + x2) = – log (1 + y2) + log k
⇒
(1 + x2) (1 + y2) = k
False, since I.F. = e
sec xdx
elog(sec x
tan x )
y (secx + tanx) = (sec x tan x) tan xdx =
= secx + tanx, the solution is,
∫ ( sec x tan x + sec x − 1) dx
2
=
secx + tanx – x +k
(ix)
True, x + y = tan–1y ⇒ 1
⇒
dy
dx
1 dy
dy
dx 1 y 2 dx
⎛ 1
⎞
dy
– 1⎟ =1 , i.e.,
⎜
2
dx
⎝ 1+ y
⎠
(1 y 2 )
which satisfies the given equation.
y2
DIFFERENTIAL EQUATIONS 193
(x)
False, because y = x does not satisfy the given differential equation.
9.3 EXERCISE
Short Answer (S.A.)
1.
2.
3.
dy
2y x .
dx
Find the differential equation of all non vertical lines in a plane.
Find the solution of
dy
e 2 y and y = 0 when x = 5.
dx
Find the value of x when y = 3.
Given that
4.
Solve the differential equation (x2 – 1)
5.
Solve the differential equation
6.
Find the general solution of
7.
Solve the differential equation
8.
Solve: ydx – xdy = x2ydx.
9.
Solve the differential equation
dy
2 xy y
dx
dy
ay emx
dx
dy
1 ex
dx
of y
2
dy
= y.
dx
2 sin x dy
= – cosx and y (0) = 1, then find the value
1 y
dx
.
12. If y(t) is a solution of (1 + t)
y (1) = –
y
dy
= 1 + x + y2 + xy2, when y = 0, x = 0.
dx
10. Find the general solution of (x + 2y3)
11. If y(x) is a solution of
1
dy
+ 2xy = 2
.
x 1
dx
1
.
2
dy
– ty = 1 and y (0) = – 1, then show that
dt
194 MATHEMATICS
13. Form the differential equation having y = (sin–1x)2 + Acos–1x + B, where A and B
are arbitrary constants, as its general solution.
14. Form the differential equation of all circles which pass through origin and whose
centres lie on y-axis.
15. Find the equation of a curve passing through origin and satisfying the differential
2
equation (1 x )
16. Solve : x2
dy
2 xy 4 x 2 .
dx
dy
= x2 + xy + y2.
dx
17. Find the general solution of the differential equation (1 + y2) + (x – etan–1y)
dy
= 0.
dx
18. Find the general solution of y2dx + (x2 – xy + y2) dy = 0.
19. Solve : (x + y) (dx – dy) = dx + dy.[Hint: Substitute x + y = z after seperating dx
and dy]
dy
= 0, given that y (1) = – 2.
dx
21. Solve the differential equation dy = cosx (2 – y cosecx) dx given that y = 2 when
20. Solve : 2 (y + 3) – xy
π
.
2
22. Form the differential equation by eliminating A and B in Ax2 + By2 = 1.
23. Solve the differential equation (1 + y2) tan–1x dx + 2y (1 + x2) dy = 0.
24. Find the differential equation of system of concentric circles with centre (1, 2).
x=
Long Answer (L.A.)
d
( xy ) = x (sinx + logx)
dx
26. Find the general solution of (1 + tany) (dx – dy) + 2xdy = 0.
25. Solve : y +
27. Solve :
dy
= cos(x + y) + sin (x + y).[Hint: Substitute x + y = z]
dx
dy
3 y sin 2 x .
dx
29. Find the equation of a curve passing through (2, 1) if the slope of the tangent to
28. Find the general solution of
x2 y2
the curve at any point (x, y) is
.
2 xy
DIFFERENTIAL EQUATIONS 195
30. Find the equation of the curve through the point (1, 0) if the slope of the tangent
to the curve at any point (x, y) is
y 1
.
x2 x
31. Find the equation of a curve passing through origin if the slope of the tangent to
the curve at any point (x, y) is equal to the square of the difference of the abcissa
and ordinate of the point.
32. Find the equation of a curve passing through the point (1, 1). If the tangent
drawn at any point P (x, y) on the curve meets the co-ordinate axes at A and B
such that P is the mid-point of AB.
dy
y (log y – log x + 1)
dx
Objective Type
Choose the correct answer from the given four options in each of the Exercises from
34 to 75 (M.C.Q)
33. Solve : x
d2y
dx 2
34. The degree of the differential equation
(A) 1
(B) 2
(C) 3
35. The degree of the differential equation 1
(A) 4
2
(B)
3
2
dy
dx
2
x sin
(D) not defined
dy
dx
3
2 2
d2y
is
dx 2
(C) not defined
d2y
36. The order and degree of the differential equation
dx 2
respectively, are
(A) 2 and not defined
(B) 2 and 2
(C) 2 and 3
37. If y = e–x (Acosx + Bsinx), then y is a solution of
d2y
dy
2
0
(A)
2
dx
dx
(C)
d2y
dy
2
2y 0
2
dx
dx
dy
is:
dx
d2y
dy
−2 + 2y = 0
(B)
2
dx
dx
(D)
d2y
+ 2y =0
dx 2
(D) 2
dy
dx
1
4
1
+ x5
(D) 3 and 3
0,
196 MATHEMATICS
38. The differential equation for y = Acos αx + Bsin αx, where A and B are arbitrary
constants is
d2y
(A)
dx 2
2
d2y
(B)
dx 2
y 0
2
y 0
d2y
d2y
y
0
y 0
(C)
(D)
dx 2
dx 2
39. Solution of differential equation xdy – ydx = 0 represents :
(A) a rectangular hyperbola
(B) parabola whose vertex is at origin
(C) straight line passing through origin
(D) a circle whose centre is at origin
40. Integrating factor of the differential equation cosx
(A) cosx
(B) tanx
(C) secx
dy
+ ysinx = 1 is :
dx
(D) sinx
41. Solution of the differential equation tany sec2x dx + tanx sec2ydy = 0 is :
(A) tanx + tany = k
(B) tanx – tany = k
tan x
k
tan y
(D) tanx . tany = k
(C)
42. Family y = Ax + A3 of curves is represented by the differential equation of degree
:
(A) 1
(B) 2
43. Integrating factor of
(A) x
44. Solution of
(A) xy
(C) 3
(D) 4
xdy
– y = x4 – 3x is :
dx
(B) logx
(C)
1
x
dy
y 1 , y (0) = 1 is given by
dx
= – ex
(B) xy = – e–x (C) xy
(D) – x
= – 1 (D) y = 2 ex – 1
DIFFERENTIAL EQUATIONS 197
dy y +1
45. The number of solutions of dx = x −1 when y (1) = 2 is :
(A) none
(B) one
(C) two
(D) infinite
46. Which of the following is a second order differential equation?
(A) (y′)2 + x = y2
(B) y′y′′ + y = sinx
(C) y′′′ + (y′′)2 + y = 0
(D) y′ = y2
47. Integrating factor of the differential equation (1 – x2)
x
1 x2
dy
− xy =1 is
dx
1
log (1 – x2)
2
48. tan–1 x + tan–1 y = c is the general solution of the differential equation:
(A) – x
(B)
(C) 1 x 2
(D)
dy 1 + y 2
=
(A)
dx 1 + x 2
dy 1 + x 2
=
(B)
dx 1 + y 2
(C) (1 + x2) dy + (1 + y2) dx = 0
(D) (1 + x2) dx + (1 + y2) dy = 0
dy
+ x = c represents :
dx
(A) Family of hyperbolas
(B) Family of parabolas
(C) Family of ellipses
(D) Family of circles
50. The general solution of ex cosy dx – ex siny dy = 0 is :
(A) ex cosy = k
(B) ex siny = k
49. The differential equation y
(C) ex = k cosy
(D) ex = k siny
3
d 2 y ⎛ dy ⎞
+ ⎜ ⎟ + 6 y 5 = 0 is :
51. The degree of the differential equation
2
dx
⎝ dx ⎠
(A) 1
(B) 2
(C) 3
(D) 5
dy
+ y = e – x , y (0) = 0 is :
dx
(A) y = ex (x – 1)
(B) y = xe–x
–x
(C) y = xe + 1
(D) y = (x + 1)e–x
52. The solution of
198 MATHEMATICS
dy
y tan x – sec x 0 is:
dx
(B) secx
(D) esecx
53. Integrating factor of the differential equation
(A) cosx
(C) ecosx
54. The solution of the differential equation
(A) y = tan–1x
(C) x = tan–1y
dy 1 y 2
is:
dx 1 x 2
(B) y – x = k (1 + xy)
(D) tan (xy) = k
55. The integrating factor of the differential equation
(A)
x
ex
(B)
dy
1+ y
+y =
is:
dx
x
ex
x
(C) xex
(D) ex
56. y = aemx + be–mx satisfies which of the following differential equation?
(A)
dy
my 0
dx
(B)
dy
my 0
dx
(C)
d2y
m2 y 0
dx 2
(D)
d2y
m2 y 0
dx 2
57. The solution of the differential equation cosx siny dx + sinx cosy dy = 0 is :
(A)
sin x
c
sin y
(B) sinx siny = c
(C) sinx + siny = c
58. The solution of x
ex
(A) y =
x
x
k
x
(C) y = xe + k
(D) cosx cosy = c
dy
+ y = ex is:
dx
(B) y = xex + cx
ey
(D) x =
y
k
y
DIFFERENTIAL EQUATIONS 199
59. The differential equation of the family of curves x2 + y2 – 2ay = 0, where a is
arbitrary constant, is:
(A) (x2 – y2)
dy
= 2xy
dx
(C) 2 (x2 – y2)
(B) 2 (x2 + y2)
dy
= xy
dx
(D) (x2 + y2)
dy
= xy
dx
dy
= 2xy
dx
60. Family y = Ax + A3 of curves will correspond to a differential equation of order
(A) 3
(B) 2
61. The general solution of
(A) e x
2
−y
(C) 1
(D) not defined
dy
2
= 2x e x − y is :
dx
2
(B) e–y + e x = c
=c
2
(C) ey = e x + c
(D) e x
2
+y
=c
62. The curve for which the slope of the tangent at any point is equal to the ratio of
the abcissa to the ordinate of the point is :
(A) an ellipse
(B) parabola
(C) circle
(D) rectangular hyperbola
x2
dy
63. The general solution of the differential equation
e 2 + xy is :
dx
(A) y ce
x2
2
x2
(B) y ce 2
x2
x2
(C) y = ( x + c) e 2
(D) y (c x)e 2
64. The solution of the equation (2y – 1) dx – (2x + 3)dy = 0 is :
2x 1
(A) 2 y 3 k
2 y +1
(B) 2 x − 3 = k
2x 3
(C) 2 y 1 k
2x 1
(D) 2 y 1 k
200 MATHEMATICS
65. The differential equation for which y = acosx + bsinx is a solution, is :
d2y
(A)
+y=0
dx 2
d2y
(B)
–y=0
dx 2
d2y
+ (a + b) y = 0
(C)
dx 2
d2y
(D)
+ (a – b) y = 0
dx 2
dy
+ y = e–x, y (0) = 0 is :
dx
(B) y = xex
(A) y = e–x (x – 1)
66. The solution of
(C) y = xe–x + 1
(D) y = xe–x
67. The order and degree of the differential equation
d3y
dx3
2
d2y
dy
3 2 2
dx
dx
(A) 1, 4
4
(B) 3, 4
y 4 are :
(C) 2, 4
(D) 3, 2
⎡ ⎛ dy ⎞ 2 ⎤ d 2 y
68. The order and degree of the differential equation ⎢1 + ⎜ dx ⎟ ⎥ = dx 2 are :
⎢⎣ ⎝ ⎠ ⎥⎦
3
(B) 2, 3
(C) 2, 1
(D) 3, 4
2
69. The differential equation of the family of curves y2 = 4a (x + a) is :
(A) 2,
2
(A) y = 4
d2y
(C) y 2
dx
dy ⎛
dy ⎞
⎜ x+ ⎟
dx ⎝
dx ⎠
dy
dx
2
(B) 2 y
dy
4a
dx
2
0
dy
⎛ dy ⎞
+ y⎜ ⎟ – y
(D) 2 x
dx
⎝ dx ⎠
70. Which of the following is the general solution of
(A) y = (Ax + B)ex
(C) y = Aex + Be–x
d2y
dy
−2
+ y = 0?
2
dx
dx
(B) y = (Ax + B)e–x
(D) y = Acosx + Bsinx
DIFFERENTIAL EQUATIONS 201
dy
+ y tan x = sec x is :
dx
(A) y secx = tanx + c
(B) y tanx = secx + c
71. General solution of
(C) tanx = y tanx + c
(D) x secx = tany + c
dy y
sin x is :
dx x
(B) x (y – cosx) = sinx + c
(D) x (y + cosx) = cosx + c
72. Solution of the differential equation
(A) x (y + cosx) = sinx + c
(C) xy cosx = sinx + c
73. The general solution of the differential equation (ex + 1) ydy = (y + 1) exdx is:
(A) (y + 1) = k (ex + 1)
(B) y + 1 = ex + 1 + k
ex 1
y
log
(D)
y 1
x
(C) y = log {k (y + 1) (e + 1)}
74. The solution of the differential equation
dy
= ex–y + x2 e–y is :
dx
(A) y = ex–y – x2 e–y + c
(C) ex + ey =
k
x3
+c
3
(B) ey – ex =
x3
+c
3
(D) ex – ey =
x3
+c
3
dy 2 xy
75. The solution of the differential equation dx
1 x2
1
is :
(1 x 2 )2
y
= c + tan–1x
1 x2
(A) y (1 + x2) = c + tan–1x
(B)
(C) y log (1 + x2) = c + tan–1x
(D) y (1 + x2) = c + sin–1x
76. Fill in the blanks of the following (i to xi)
dy
(i)
d 2 y dx
0 is _________.
e
The degree of the differential equation
dx 2
(ii)
The degree of the differential equation 1
dy
dx
2
x is _________.
202 MATHEMATICS
(iii)
The number of arbitrary constants in the general solution of a differential
equation of order three is _________.
(iv)
dy
dx
y
x log x
1
is an equation of the type _________.
x
(v)
dx
General solution of the differential equation of the type dy + P1 x = Q1
is given by _________.
77.
xdy
2 y x 2 is _________.
dx
(vi)
The solution of the differential equation
(vii)
The solution of (1 + x2)
(viii)
The solution of the differential equation ydx + (x + xy)dy = 0 is ______.
(ix)
General solution of
(x)
The solution of differential equation coty dx = xdy is _________.
(xi)
The integrating factor of
dy
+2xy – 4x2 = 0 is _________.
dx
dy
dx
y = sinx is _________.
dy
1 y
y
is _________.
dx
x
State True or False for the following:
(i)
dx
Integrating factor of the differential of the form dy + p1 x = Q1 is given
by e ∫ p1dy .
(ii)
dx
Solution of the differential equation of the type dy + p1 x = Q1 is given
by x.I.F. = (I.F) Q1dy .
(iii)
Correct substitution for the solution of the differential equation of the
dy
f ( x, y ) , where f (x, y) is a homogeneous function of zero
dx
degree is y = vx.
type
DIFFERENTIAL EQUATIONS 203
(iv)
Correct substitution for the solution of the differential equation of the
type
dx
g ( x, y ) where g (x, y) is a homogeneous function of the
dy
degree zero is x = vy.
(v)
Number of arbitrary constants in the particular solution of a differential
equation of order two is two.
(vi)
The differential equation representing the family of circles
x2 + (y – a)2 = a2 will be of order two.
y
x
1
3
(vii)
dy
The solution of
dx
(viii)
Differential equation representing the family of curves
y = ex (Acosx + Bsinx) is
2
2
is y 3 – x 3 = c.
d2y
dy
–2
2y 0
2
dx
dx
dy x + 2 y
=
is x + y = kx2.
dx
x
(ix)
The solution of the differential equation
(x)
Solution of
(xi)
The differential equation of all non horizontal lines in a plane is
d2x
=0 .
dy 2
xdy
dx
y x tan
y
y
is sin
x
x
cx